To find the solution to the linear-quadratic system given by the equations:
- \( y = x^2 + 3x - 13 \)
- \( y = 5 \)
We can substitute \( y = 5 \) into the first equation to find the corresponding \( x \)-values.
Substituting \( y = 5 \):
\[ 5 = x^2 + 3x - 13 \]
Rearranging this gives us:
\[ x^2 + 3x - 18 = 0 \]
Next, we can factor this quadratic equation:
\[ (x + 6)(x - 3) = 0 \]
Setting each factor to zero gives us:
\[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \] \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
The solutions for \( x \) are \( x = -6 \) and \( x = 3 \). Since \( y = 5 \), we can now write the solutions as points in the coordinate system:
- \( (-6, 5) \)
- \( (3, 5) \)
Now, looking at your choices, the correct response is:
(3, 5) and (-6, 5)